最大子序列和四种求解方法

1.暴力求解：时间复杂度O(N^3)

int maxSubSum1(const vector<int> & a)
{
    
    
    int maxSum = 0;
    for (int i = 0; i < a.size(); ++i)
        for (int j = i; j < a.size(); ++j) {
    
    
            int thisSum = 0;
            for (int k = i; k <= j; ++k)
                thisSum += a[k];
            if (thisSum > maxSum)
                maxSum = thisSum;
        }
    return maxSum;
}

2.改进的暴力求解：时间复杂度O(N^2)

int maxSubSum2(const vector<int> & a)
{
    
    
    int maxSum = 0;
    for (int i = 0; i < a.size(); ++i) {
    
    
        int thisSum = 0;
        for (int j = i; j < a.size(); ++j) {
    
    
            thisSum += a[j];
            if (thisSum > maxSum)
                maxSum = thisSum;
        }
    }
    return maxSum;
}

3.分治：时间复杂度O(N logN)

int max3(int a, int b, int c)
{
    
    
    int res = 0;
    res = a > b ? a : b;
    res = res > c ? res : c;
    return res;
}

int maxSumRec(const vector<int> & a, int left, int right)
{
    
    
    if (left == right)
        if (a[left] > 0)
            return a[left];
        else
            return 0;

    int center = (left + right) / 2;
    int maxLeftSum = maxSumRec(a, left, center);
    int maxRightSum = maxSumRec(a, center + 1, right);
    int maxLeftBorderSum = 0, leftBorderSum = 0;
    for (int i = center; i >= left; --i) {
    
    
        leftBorderSum += a[i];
        if (leftBorderSum > maxLeftBorderSum)
            maxLeftBorderSum = leftBorderSum;
    }
    int maxRightBorderSum = 0, rightBorderSum = 0;
    for (int j = center + 1; j <= right; ++j) {
    
    
        rightBorderSum += a[j];
        if (rightBorderSum > maxRightBorderSum)
            maxRightBorderSum = rightBorderSum;
    }
    return max3(maxLeftSum, maxRightSum, maxLeftBorderSum + maxRightBorderSum);
}

int maxSubSum3(const vector<int> & a)
{
    
    
    return maxSumRec(a, 0, a.size() - 1);
}

4.联机算法：时间复杂度O(N)

int maxSubSum4(const vector<int> & a)
{
    
    
    int thisSum = 0, maxSum = 0;
    for (int j = 0; j < a.size(); ++j) {
    
    
        thisSum += a[j];
        if (thisSum > maxSum)
            maxSum = thisSum;
        else if (thisSum < 0)
            thisSum = 0;
    }
    return maxSum;
}